If $f\in C^{\infty}$ has infinitely many zeros and if the $f^{(n)}$ is $\mathcal{O}(n!)$, then $f$ is constant on some interval The question goes as follows:

Let $f\in C^{\infty}((-2,2))$ and suppose $f(x)=0$ for infinitely many
$x$'s in $[-1,1]$. Suppose also that there exists a constant $C>0$
such that, for sufficiently large $n$, we have
$\|f^{(n)}\|_{\mathrm{sup}}<Cn!$ (so $\|f^{(n)}\|_{\mathrm{sup}} 
 = \mathcal{O}(n!)$). Prove that there exists an open subinterval of
$(-2,2)$ on which $f$ vanishes.

My attempt so far:
$\bullet$ Since $f(x)=0$ for infinitely many $x$'s in $[-1,1]$, there exists an accumulation point $x_0$ among these $x$'s. My intuition is that the open interval we seek will contain this $x_0$.
$\bullet$ The $\mathcal{O}(n!)$ condition reminds me of Taylor expansion. To this end, suppose for some $n$, $f^{(m)}$ is bounded by $Cm!$ for all $m\geqslant n$. We can try to expand  $f$ around $x_0$ using
$$
  f(x) = f(x_0) + \sum_{j=1}^{n-1} \frac{f^{(j)}(x_0)}{j!}(x-x_0)^{j} + \sum_{k=n}^{\infty} \frac{f^{(k)}(x_0)}{k!} (x-x_0)^{k},
$$
where the first term is $0$ and the last is bounded by $C (x-x_0)^{n} /(1-(x-x_0))$. I'm not sure if this is useful though, since in this case $\lvert x-x_0\rvert $ would depend on the choice of $\epsilon$.
$\bullet$ Maybe I should try to prove this by contradiction? For example suppose WLOG that there exists a sequence $y_n\to x$ with $y_n>x$ such that $y_n-x < 1 /n$ and $f(y_n) > 0$ for all $n$?
Any hints would be greatly appreciated.
 A: I have the following rough idea.
As $[-1,1]$ is compact, we have an accumulation point of the zeros in $[-1,1]$. Call it $x_0$. By continuity, we have $f(x_0)=0$.
Since in every open interval $I$ containing $x_0$, there exists a zero other than $x_0$, by the mean value theorem, $I$ also contains a zero of $f'$. This means that $x_0$ is also the accumulation point of the zeros of $f'$. By continuity of $f'$, we have $f'(x_0)=0$.
Repeating the same reasoning, we have that $f^{(n)}(x_0)=0$ for any $n\in\mathbb{N}\cup\{0\}$. Then applying the Taylor's theorem, with the remainder term taken to be of the Lagrange form, we have for any $y\in [-1,1]$ and any $n\in\mathbb{N}\setminus\{0\}$, there exists some $\xi_n$ in the interval with endpoints $x_0$ and $y$ such that
$$|f(y)|=\left|\frac{f^{(n)}(\xi_n)}{n!}(x_0-y)^n\right|\leq C|x_0-y|^n.$$
Taking the limit gives that $f(y)=0$.
If you find any problems with the arguments above, please let me know. Thank you!

Update: Mr./Ms. Skorpion pointed out that $y$ should satisfy $|x_0-y|<1$ so that $\lim_{n\rightarrow+\infty}|x_0-y|^n=0$.
